Prove that (sinx / 1+cos x) + (1-cos x / sin x) = 2sin x / 1+cos x?

Question

i've looked at this question for half an hour, and still have no idea how to answer it. just a clue on what to do would be appreciated.
thanks

Answer 1

The 1st term on the LHS is a fixed multiple of RHS so do not disturb the 1st term

now convert the second term (1-cos x)/sin x to (1+cos x) in denominator.

how do we do it
multiply by (1+cos x)/(1+cos x)
now numerator = (1-cos x)(1+cos x) = 1- cos ^2 x = sin^2 x
denominator = sin x(1+cos x)
so value = sin x/(1+cos x)
now LHS = sin x/(1+cos x) + sin x/(1+cos x) = 2 sin x /(1+cos x)
= rhs
simple is it not ?

Answer 2

Get the LCM,
i.e.
(sin x) / (1+cos x) + (1-cos x) / (sin x) =
[sin^2 x +1 - cos^2 x] / (sin x).(1+cos x) =
(1-cos 2x)/[(sin x).(1+cos x)]

because cos^2 x - sin^2 x = cos 2x,
this is then equal to
2sin^2 x / [(sin x).(1+cos x)]
now one sine goes off and you get
2sin x / (1+cos x)

the above is because
cos 2x = 1 - 2sin^2 x => 2sin^2 x = 1-cos 2x , as had appeared on the numerator.
QED then, hope this helps!

Answer 3

First, put in parentheses in your denominator. Now put everything over a common denominator. In the top of the first term you will have sin^2 x and in the top of the second you will have (1-cos x)(1+cos x)=1-cos^2 x=sin^2 x. Now add and cancel a sin x.

Answer 4

sin x/(1+cos x)+(1-cos x)/sin x = 2 sin x/(1+cos x)

let us prove the right hand side of the identity so we will transform the left hand side...

sin x/(1+cos x)+(1-cos x)/sin x

take the LCD which is sin x(1+cos x)...

[sin^2 x + (1-cos^2 x)]/sin x(1+cos x)

Since 1-cos^2 x is equal to sin^2 x from the identity sin^2 x+cos^2 x=1 then...

(sin^2 x + sin^2 x)/ sinx(1+cos x)

Add like terms sin^2 x...

2 sin^2 x/ sin x(1+cos x)

Simplifying the expression since sin^2 x/sin x=sin x

2 sin x/(1+cos x)

therefore we have proved that the left hand side expression is equal to the right hand expression...

cheers

Answer 5

Start by multiplying both sides by (1+cos(x)). Then subtract sin(x) from both sides. Multiply both sides by sin(x) and you'll have 1-cos²(x) = sin²(x) and therefore
1 = sin²(x) + cos²(x) which you should have had as one of your very first trig identities.


Doug

Answer 6

sinx/(1+cosx) + (1-cosx)/sinx =

Multiplying the second term by (1+cosx)/(1+cosx), we get

sinx/(1+cosx)+
(1-cosx)(1+cosx)/(1+cosx)sinx=
sinx/((1+cosx)+
(1-cos^2x)/(1+cosx)sinx

But

sin^2x+cos^2x=1 (A known identity in trigonometry), and
sin^2x=1-cos^2x

Substituting sin^2x for (1-cos^2x) in the above expression, we get

sinx/(1+cosx)
+sin^2x /(1+cosx)/sinx=
sinx/(1+cosx) +sinx/(1+cosx)=
2sinx/(1+cosx).

The key in solving the problem is in knowing that

sin^2x+cos^2x=1, and therefore
sin^2x=1-cos^2x.

This should help you in solving similar problems later.

Answer 7

cos(2x)=cos(x)*cos(x) - sin(x)*sin(x) cos(2x)=cos(x)^2 - sin(x)^2 cos(x)^2 + sin(x)^2 = a million cos(x)^2 = a million - sin(x)^2 cos(x)^2 - sin(x)^2 = a million - sin(x)^2 - sin(x)^2 cos(x)^2 - sin(x)^2 = a million - (2*sin(x)^2 )

Answer 8

good one math_kp
*salute*

math_kp solves it nice and easyy